Question

the number of real solutins of the equation -2-x^2+x=2^x

Answer 1

*solutions

Answer 2

x=2

Answer 3

Use the discriminant "\[b^2-4ac\]" if this is less than zero then there are no real roots, if greater than zero then there are 2 real roots , if x=0 , then there 1 real root. where b= the numer of x terms, in this case that is -1, a= the number of x^2 terms, in this case 3, and c= the number constants in this case 2.

Answer 4

sorry i miss read the question, don't listen to me.

Answer 5

the answer is that there is no real solution. i dont know how to figure that out

Answer 6

so whats the answer?

Answer 7

no real solution

Answer 8

how about x=2 isn't it correct?

Answer 9

if x=2 it gives the equality of that equation
-2-2^2+2=2^2
or -2^2=2^2
or 4=4

Answer 10

no
-2^2=2^2 => -4=4 which is absurd

Answer 11

what is the rule in multiplying two (-)number ?

Answer 12

its not (-2)^2 its -(2^2)

Answer 13

so that is not an exponent?

Answer 14

what are you talking about?

Answer 15

ok shall we say 3^2
so 2 is the exponent of 3.

Answer 16

did this question ever get answered?

Answer 17

It does not appear that it did...
the number of real solutins of the equation \[-2-x^2+x=2^x\]

Answer 18

how did you come up that equation?

Answer 19

I copy and pasted from the original...

Answer 20

so whats your final answer?

Answer 21

I agree with the given, "no real solution"

Answer 22

ok . am,what is real solution all about?

Answer 23

my argument is that on the L.H.S. you have a polynomial degree 2 and on the R.H.S.
you have the exponential expression \[2^x\]

\[2^x\] is always positive and increasing

so now we look at the L.H.S. more closely

since it is a polynomial of degree 2, the highest degree term i.e. \[-x^2\] will dominate and thus the L.H.S. will be negative for "large" values of "x"

now we observe that the quantity \[-x^2+x\] will only be positive when \[x\in(0,1)\]
and moreover will never be greater than 2

Therefore the L.H.S. is always negative but the R.H.S' is always positve.
Hence no real number solution
i.e. if a solution does in fact exist to this equation it is not a real number value...

Answer 24

Rewrite the equation as
\(x^2-x+2+2^x=0\)
Notice that
\(x^2-x+2+2^x=(x-\frac{1}{2})^2+\frac{7}{4}+2^x >0\)
So obviously the equation has no solution.

Answer 25

Clearly.

Answer 26

what exactly does it mean when it is said the an expression has a real solution. i mean whit is this "solution" all about?